O is the circumcentre of a triangle `DeltaABC`. The point A and the chord BC are on the opposite side of O. If `/_BOC = 150^@`. Then the angle `/_BAC `is :

Options

A. `65^@`
B. `60^@`
C. `70^@`
D. `75^@`

Answer

To solve for the angle \( \angle BAC \) in triangle \( \Delta ABC \) where \( O \) is the circumcenter and \( \angle BOC = 150^\circ \), we can follow these steps:

### Step 1: Understand the relationship between the angles
In a triangle, the angle at the circumcenter \( O \) (which is \( \angle BOC \)) is related to the angle at the opposite vertex \( A \) (which is \( \angle BAC \)). Specifically, the angle at the circumcenter is twice the angle at the vertex opposite to the chord.

### Step 2: Apply the relationship
Given that \( \angle BOC = 150^\circ \), we can use the relationship:
\[ \angle BAC = \frac{1}{2} \angle BOC \]

### Step 3: Substitute the known value
Now, substitute the value of \( \angle BOC \) into the equation:
\[ \angle BAC = \frac{1}{2} \times 150^\circ \]

### Step 4: Calculate the angle
Now perform the calculation:
\[ \angle BAC = \frac{150^\circ}{2} = 75^\circ \]

### Conclusion
Thus, the angle \( \angle BAC \) is \( 75^\circ \).

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